## Precalculus (6th Edition)

False Correct Statement: $(x+y)^{-1} = \dfrac{1}{x+y}$
RECALL: $a^{-m} = \dfrac{1}{a^m}$ Use the rule above to obtain: $(x+y)^{-1} = \dfrac{1}{(x+y)^1}=\dfrac{1}{x+y}$ Thus, the given statement is false. The correct statement is: $(x+y)^{-1} = \dfrac{1}{x+y}$